to make the two polynomi(als equivalent, then multiply each term by $(a^pr-br)$
so that
$$(a^pr)^n-(br)^n=br^{n-1}(a^pr-br)+br^{n-2}a^p(a^pr-br)+br^{n-3}a^{2p}(a^pr-br)+\cdots +bra^{p(n-2)}(a^pr-br)+a^{p(n-1)}(a^pr-br)$$

since two polynomials are regular and have equal number of terms, then the first terms in both are equal, so are the rest corresponding terms.
we wont mind if $(a^pr-br)$ is $b^w$.
The fact is, $br^{n-1}=br^{n-1}(a^pr-br)$

Also, if $(a^pr-br)=b^w$ then the $(a^pr)^n-(br)^n$ would not be as it is but would be $(a^pr)^{n+w}-(br)^{n+w}$. That's the only way it would hold.

Let $x_n$ be an odd positive integer. From the sequence’s formula, $x_{n+1}=\frac{3x_n+1}{2^k}$, $x_{n+2}=\frac{3x_{n+1}+1}{2^m}$ and so forth, $m,k\in \mathbb{Z^+}$.\\
For there to exist a cycle in the sequence, there must exist an odd integer $x_0$ such that $x_n=x_0$.\\
Let $x_1$, the next odd integer after $x_0$, be given by
$$x_1=\frac{3x_0+1}{2^{a_1}}$$
$x_2$ will be given by
$$x_2=\frac{3x_1+1}{2^{a_2}}$$
$x_2$ in terms of $x_0$ will be
\begin{align*}
x_2&=\frac{3(\frac{3x_0+1}{2^{a_1}})+1}{2^{a_2}}\\
&=\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}}
\end{align*}
$x_3$ will be given by
\begin{align*}
x_3&=\frac{3x_2+1}{2^{a_3}}\\
&=\frac{3(\frac{3x_1+1}{2^{a_2}})+1}{2^{a_3}}\\
&=\frac{3(\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}})+1}{2 ^{a_3}}\\
&=\frac{27x_0+9+3.2^{a_1}+2^{a_1+a_2}}{2^{a_1+a_2+ a_3}}
\end{align*}
From the three examples, we can generate a formula for $x_n$ in terms of $x_0$, which will be
$$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$
Theorem
If $x_n=x_0$, then $x_n=1$.
Proof
Let $x_n=x_0$, hence
$$x_0=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$
Let $\sum\limits_{n=1}^na_i$ be $k$
$$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$
$2^k$ can be expressed as $(2^{\frac{k}{n}})^n$ and $x_0$ as $(\sqrt[n]{x_0})^n$
$$\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}\tag{1}$$
When factoring,
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})$$
This is same as
$$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1})$$
We therefore notice equation (1) holds iff the polynomial is regular.
We factorise $\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n$.
\begin{align*}
\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\ (\{3\!\sqrt[n]{x_0}\}^{n-1}+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}\\
&\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2+\cdots +\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}\\
&\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1})
\\
\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
&\quad+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
&\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})+\cdots\\
&\quad+\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
&\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0}))\tag{2}
\end{align*}
Thus, each term in the polynomial in equation (1) is equal to each corresponding term in the polynomial in equation (2), for example the first term in polynomial (1) is equal to either the first or last term in polynomial (2). The second term in polynomial (1) is equal to either the second or the second last term in polynomial (2) and so forth.
If the first term in polynomial (1) equals the last term in polynomial (2), then
\begin{align*}
3^{n-1}&=\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
\frac{3^{n-1}}{(2^{\frac{k}{n}})^{n-1}}&=(\sqrt[n]{x_0})^{n-1}\sqrt[n]{x_0}(2^{\frac{k}{n}}-3)\\
(\frac{3}{2^{\frac{k}{n}}})^{n-1}&=x_0(2^{\frac{k}{n}}-3)\\
x_0&=(\frac{3}{2^{\frac{k}{n}}})^{n-1}\frac{1}{(2^{\frac{k}{n}}-3)}
\end{align*}
But $2^{\frac{k}{n}}>3$. Hence, $x_0<1$ in this case.

If the first term in polynomial (1) equals the first term in polynomial (2), then
\begin{align*}
3^{n-1}&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
1&=(\sqrt[n]{x_0})^{n-1}\sqrt[n]{x_0}(2^{\frac{k}{n}}-3)\\
x_0&=\frac{1}{(2^{\frac{k}{n}}-3)}\tag{3}
\end{align*}
The next pair of equal terms,
\begin{align*}
3^{n-2}2^{a_1}&=\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
2^{a_1}&=\sqrt[n]{x_0}^{n-2}\sqrt[n]{x_0}\sqrt[n]{x_0}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\\
2^{a_1}&=x_02^{\frac{k}{n}}(2^{\frac{k}{n}}-3)
\end{align*}
Substituting (3)
\begin{align*}
2^{a_1}&=\frac{1}{(2^{\frac{k}{n}}-3)}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\\
2^{a_1}&=2^{\frac{k}{n}}
\end{align*}
Substituting $2^{a_1}$ in equation (3),
$$x_0=\frac{1}{(2^{a_1}-3)}$$
The only value that satisfies this equation such that $x_0$ and $a_1$ are both positive integers is $a_1=2$ to give $x_0=1$.

Thus, each term in the polynomial in equation (1) is equal to each corresponding term in the polynomial in equation (2), for example the first term in polynomial (1) is equal to either the first or last term in polynomial (2). The second term in polynomial (1) is equal to either the second or the second last term in polynomial (2) and so forth.
If the first term in polynomial (1) equals the last term in polynomial (2)

Let $x_n$ be an odd positive integer. From the sequence’s formula, $x_{n+1}=\frac{3x_n+1}{2^k}$, $x_{n+2}=\frac{3x_{n+1}+1}{2^m}$ and so forth, $m,k\in \mathbb{Z^+}$.\\
For there to exist a cycle in the sequence, there must exist an odd integer $x_0$ such that $x_n=x_0$.\\
Let $x_1$, the next odd integer after $x_0$, be given by
$$x_1=\frac{3x_0+1}{2^{a_1}}$$
$x_2$ will be given by
$$x_2=\frac{3x_1+1}{2^{a_2}}$$
$x_2$ in terms of $x_0$ will be
\begin{align*}
x_2&=\frac{3(\frac{3x_0+1}{2^{a_1}})+1}{2^{a_2}}\\
&=\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}}
\end{align*}
$x_3$ will be given by
\begin{align*}
x_3&=\frac{3x_2+1}{2^{a_3}}\\
&=\frac{3(\frac{3x_1+1}{2^{a_2}})+1}{2^{a_3}}\\
&=\frac{3(\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}})+1}{2 ^{a_3}}\\
&=\frac{27x_0+9+3.2^{a_1}+2^{a_1+a_2}}{2^{a_1+a_2+ a_3}}
\end{align*}
From the three examples, we can generate a formula for $x_n$ in terms of $x_0$, which will be
$$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$
Theorem
If $x_n=x_0$, then $x_n=1$.
Proof
Let $x_n=x_0$, hence
$$x_0=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$
Let $\sum\limits_{n=1}^na_i$ be $k$
$$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$
$2^k$ can be expressed as $(2^{\frac{k}{n}})^n$ and $x_0$ as $(\sqrt[n]{x_0})^n$
$$\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}\tag{1}$$
When factoring,
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})$$
This is same as
$$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1})$$
We therefore notice equation (1) holds iff the polynomial is regular.
We factorise $\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n$.
\begin{align*}
\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\ (\{3\!\sqrt[n]{x_0}\}^{n-1}+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}\\
&\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2+\cdots +\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}\\
&\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1})
\\
\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
&\quad+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
&\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})+\cdots\\
&\quad+\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
&\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0}))\tag{2}
\end{align*}
Thus, the number of terms in the two polynomials is equal. The first term in the polynomial in equation (1) is equal to any term in the polynomial in equation (2). Each term in polynomial 2 can be defined by the following,
$$3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$
Where $r=1,2,3,...,n$
Hence,
\begin{align*}
3^{n-1}&=3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
3^{n-1}&=3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)\\
x_0&=\frac{3^{n-1}}{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}(2^{\frac{k}{n}}-3)}\\
x_0&=\{\frac{3}{2^{\frac{k}{n}}}\}^{r-1}\frac{1}{(2^{\frac{k}{n}}-3)}
\end{align*}
But $2^{\frac{k}{n}}>3$. Hence, $x_0<1$ for $r>1$.
When $r=1$,
\begin{align*}
3^{n-1}&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
1&=(\sqrt[n]{x_0})^{n-1}\sqrt[n]{x_0}(2^{\frac{k}{n}}-3)\\
x_0&=\frac{1}{(2^{\frac{k}{n}}-3)}\tag{3}
\end{align*}
The next term in the polynomial in equation (1), is equal to any term in polynomial in equation (2), except $\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$. Therefore,
$$3^{n-2}2^{a_1}=\{3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$
Such that $r=2,3,4,...,n$
\begin{align*}
3^{n-2}2^{a_1}&=3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)\\
2^{a_1}&=\frac{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)}{3^{n-2}}
\end{align*}
We substitute (3)
\begin{align*}
2^{a_1}&=\frac{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}\frac{1}{(2^{\frac{k}{n}}-3)}(2^{\frac{k}{n}}-3)}{3^{n-2}}\\
2^{a_1}&=3^{2-r}\{2^{\frac{k}{n}}\}^{r-1}\\
\end{align*}
But $a_1$ is an integer. Therefore $r=2$ to eliminate $3^{2-r}$. Hence,
$$2^{a_1}=2^{\frac{k}{n}}$$
Substituting $2^{a_1}$ in equation (3),
$$x_0=\frac{1}{(2^{a_1}-3)}$$
The only value that satisfies this equation such that $x_0$ and $a_1$ are both positive integers is $a_1=2$ to give $x_0=1$.

So, this is what I have as the proof that there exists only one Cycle in the Collatz sequence. I haven't worked out on the rest of the proof but maybe some time in future.

Let $x_n$ be an odd positive integer. From the sequence’s formula, $x_{n+1}=\frac{3x_n+1}{2^k}$, $x_{n+2}=\frac{3x_{n+1}+1}{2^m}$ and so forth, $m,k\in \mathbb{Z^+}$.
For there to exist a cycle in the sequence, there must exist an odd integer $x_0$ such that $x_n=x_0$.
Let $x_1$, the next odd integer after $x_0$, be given by
$$x_1=\frac{3x_0+1}{2^{a_1}}$$
$x_2$ will be given by
$$x_2=\frac{3x_1+1}{2^{a_2}}$$
$x_2$ in terms of $x_0$ will be
\begin{align*}
x_2&=\frac{3(\frac{3x_0+1}{2^{a_1}})+1}{2^{a_2}}\\
&=\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}}
\end{align*}
$x_3$ will be given by
\begin{align*}
x_3&=\frac{3x_2+1}{2^{a_3}}\\
&=\frac{3(\frac{3x_1+1}{2^{a_2}})+1}{2^{a_3}}\\
&=\frac{3(\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}})+1}{2 ^{a_3}}\\
&=\frac{27x_0+9+3.2^{a_1}+2^{a_1+a_2}}{2^{a_1+a_2+ a_3}}
\end{align*}
From the three examples, we can generate a formula for $x_n$ in terms of $x_0$, which will be
$$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$

Theorem
If $x_n=x_0$, then $x_n=1$.

proof
Let $x_n=x_0$, hence
$$x_0=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$
Let $\sum\limits_{n=1}^na_i$ be $k$
$$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$
$2^k$ can be expressed as $(2^{\frac{k}{n}})^n$ and $x_0$ as $(\sqrt[n]{x_0})^n$
\begin{equation}
\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n\!=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}\!+2^{\sum\limits_{n=1}^{n-1}a_i}
\end{equation}
When factoring,
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})$$
This is same as
$$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1})$$
Therefore equation 1 holds iff the polynomial is regular.
We factorise $\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n$.
\begin{align*}
\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\ (\{3\!\sqrt[n]{x_0}\}^{n-1}+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}\\
&\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2+\cdots +\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}\\
&\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1})
\\
\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
&\quad+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
&\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})+\cdots\\
&\quad+\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
&\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\tag{2}
\end{align*}
Thus, the polynomial in equation 2 is same as the polynomial in equation 1 such that the first term in the polynomial in equation 1 is equal to any term in the polynomial in equation 2.
We define a term in the polynomial in equation 2 as
$$\{3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$
Such that $r$ is a positive integer.
We equate this to the first term in the polynomial in equation 1. Hence,
\begin{align*}
3^{n-1}&=3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
3^{n-1}&=3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)\\
x_0&=\frac{3^{n-1}}{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}(2^{\frac{k}{n}}-3)}\tag{3}
\end{align*}
We define a different term in the polynomial in equation 2 as
$$\{3\!\sqrt[n]{x_0}\}^{n-w}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{w-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$
Such that $w$ is a positive integer.
We equate this to the next term in the polynomial in equation 1.
\begin{align*}
3^{n-2}2^{a_1}&=\{3\!\sqrt[n]{x_0}\}^{n-w}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{w-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
3^{n-2}2^{a_1}&=3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}x_0(2^{\frac{k}{n}}-3)\\
x_0&=\frac{3^{n-2}2^{a_1}}{3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}(2^{\frac{k}{n}}-3)}
\end{align*}
We substitute 3
\begin{align*}
\frac{3^{n-1}}{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}(2^{\frac{k}{n}}-3)}&=\frac{3^{n-2}2^{a_1}}{3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}(2^{\frac{k}{n}}-3)}\\
\frac{3^{n-1}}{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}}&=\frac{3^{n-2}2^{a_1}}{3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}}\\
2^{a_1}&=\frac{3^{n-1}3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}}{3^{n-r}3^{n-2}\{2^{\frac{k}{n}}\}^{r-1}}\\
2^{a_1}&=3^{r-w+1}\{2^{\frac{k}{n}}\}^{w-r}\tag{4}
\end{align*}
$a_1$ is an integer. This implies the equation will hold iff $3^{r-w+1}=1$.
\begin{align*}
3^{r-w+1}&=3^0\\
r-w+1&=0\\
r&=w-1
\end{align*}
We substitute this in equation 4.
\begin{align*}
2^{a_1}&=3^{(w-1)-w+1}\{2^{\frac{k}{n}}\}^{w-(w-1)}\\
2^{a_1}&=2^{\frac{k}{n}}
\end{align*}
Substituting $2^{a_1}$ in equation 3,
\begin{align*}
x_0&=\frac{3^{n-1}}{3^{n-r}\{2^{a_1}\}^{r-1}(2^{a_1}-3)}\\
x_0&=\{\frac{3}{2^{a_1}}\}^{r-1}\frac{1}{2^{a_1}-3}
\end{align*}
$2^{a_1}=2^{\frac{k}{n}}$ and $2^{\frac{k}{n}}>3$. This implies $x_0<1$ for $r>1$. Therefore
$$x_0=\frac{1}{2^{a_1}-3}$$
The only value that satisfies this equation such that $x_0$ and $a_1$ are both positive integers is $a_1=2$ to give $x_0=1$.